(1) According to the meaning of the question, the genotype of black glume is B-, so the genotype of pure black glume parents can be expressed as BB B-；; ; And because in F2, black glume: yellow glume: white glume = 12: 3: 1 we know that the genotype of black glume in F 1 is BbYy, the genotype of white glume in F2 is bbyy, and the genotype of yellow glume is BB-, so the genotype of pure yellow glume parents can be expressed as BB-, so if we want to get F66,

(2) F2 is black glume: yellow glume: white glume = 12: 3: 1, which is a special case of 9: 3: 3: 1 and follows the law of free gene combination. Therefore, these two pairs of alleles are located on two pairs of homologous chromosomes.

(3) If F 1 crosses oat, the progeny genotypes are Bbyy, bbYy, BbYy and bbyy, and the phenotype and segregation ratio are black glume: yellow glume: white glume = 2: 1: 1.

(4) Haploid plants were obtained by anther culture in vitro. Because the offspring of Bbyy crossing with white glume plants are only BByy, and the inbred offspring will not have yellow individuals, so if the inbred offspring have yellow glume plants, the original seeds produced by black glume plants are BBYY.

So the answer is:

( 1)BByy？ bbYY

(2) Yes

(3) Heiying: Huangying: Baiying = 2: 1: 1

(4) Anthers were cultured in vitro, and Huangying plants appeared.